# Thread: Exponential radioactive decay while original amount increases.

1. ## Exponential radioactive decay while original amount increases.

This problem is much trickier than I thought.
I can solve the "easy" half-life problems, such as beginning amount, ending amount and so on. This one's tougher.

Let's suppose you have a substance "X" that has a half life of 30 days. You start out at day zero with 0 grams of "X". For each day, 1 gram of "X" is added to a beaker. (This is a continual process whereby, at the end of each hour, the amount of "X" is increased by 1/24 of a gram, or for each second the amount is increased by 1/86,400 gram. The increases are not done by "chunks".)
Okay so each day, you have an additional gram of "X" and after 5 days (for example) you have 5 grams of "X". But this isn't exactly accurate because some of that 5 grams has already started to decay with a half-life of 30 days.

So, what is the formula for determining how much "X" is present at any particular time?
And since this is my first posting - hello everyone.

2. Originally Posted by wolf
This problem is much trickier than I thought.
I can solve the "easy" half-life problems, such as beginning amount, ending amount and so on. This one's tougher.

Let's suppose you have a substance "X" that has a half life of 30 days. You start out at day zero with 0 grams of "X". For each day, 1 gram of "X" is added to a beaker. (This is a continual process whereby, at the end of each hour, the amount of "X" is increased by 1/24 of a gram, or for each second the amount is increased by 1/86,400 gram. The increases are not done by "chunks".)
Okay so each day, you have an additional gram of "X" and after 5 days (for example) you have 5 grams of "X". But this isn't exactly accurate because some of that 5 grams has already started to decay with a half-life of 30 days.

So, what is the formula for determining how much "X" is present at any particular time?
And since this is my first posting - hello everyone.
Make a function such as

$f(t)=\text{initial amount plus amount added}-\text{amount accumitavely decayed}$

3. Originally Posted by wolf
This problem is much trickier than I thought.
I can solve the "easy" half-life problems, such as beginning amount, ending amount and so on. This one's tougher.

Let's suppose you have a substance "X" that has a half life of 30 days. You start out at day zero with 0 grams of "X". For each day, 1 gram of "X" is added to a beaker. (This is a continual process whereby, at the end of each hour, the amount of "X" is increased by 1/24 of a gram, or for each second the amount is increased by 1/86,400 gram. The increases are not done by "chunks".)
Okay so each day, you have an additional gram of "X" and after 5 days (for example) you have 5 grams of "X". But this isn't exactly accurate because some of that 5 grams has already started to decay with a half-life of 30 days.

So, what is the formula for determining how much "X" is present at any particular time?
And since this is my first posting - hello everyone.

This would be a first order linear differential equation

Since the substance is not added in chunks lets suppose that is it added continously. Then the rate in would be

$r_i=1 \frac{gram}{day}$

The rate out would be

$r_0=kA$ where A is the number of grams of the substance at any time (t) and $k=\frac{\ln(2)}{30}$

$\frac{dA}{dt}=1-\frac{\ln(2)}{30}A$

$\frac{dA}{dt}+\frac{\ln(2)}{30}A=1$

Our integrating factor is $e^{\frac{\ln(2)}{30}t}$

$\frac{d}{dt}\left[e^{\frac{\ln(2)}{30}t}A \right]=e^{\frac{\ln(2)}{30}t}$

Integrating both sides and solving for A we get

$A(t)=\frac{30}{\ln(2)}+Ce^{-\frac{\ln(2)}{30}t}$

Using your inital condition that $A(0)=0$ we get $C=-\frac{30}{\ln(2)}$

$A(t)=\frac{30}{\ln(2)}-\frac{30}{\ln(2)}e^{-\frac{\ln(2)}{30}t}$

$t \to \infty \\\ A(t) \to \frac{30}{\ln(2)} \approx 43.281$

4. Wow, those replies came pretty fast.

Let's see if I can work out the calculations of the formula for 1 gram per day, for 5 days with a 30 day half life.

30/ln(2) -30/ln(2) * e^-(ln(2)/30)*5 =

43.28085 - 43.28085*e^-(ln(2)/30)*5 =

43.28085 - 43.28085*2.718281828^-((.02310491)*5) =

43.28085 - 43.28085*2.718281828^-(.11552453) =

43.28085 - 43.28085*.8908987 =

43.28085 - 38.558853 =

4.721997 grams of "X" after 5 days

That appears to be correct. Did I interpret the formula correctly?

5. Originally Posted by wolf
Wow, those replies came pretty fast.

Let's see if I can work out the calculations of the formula for 1 gram per day, for 5 days with a 30 day half life.

30/ln(2) -30/ln(2) * e^-(ln(2)/30)*5 =

43.28085 - 43.28085*e^-(ln(2)/30)*5 =

43.28085 - 43.28085*2.718281828^-((.02310491)*5) =

43.28085 - 43.28085*2.718281828^-(.11552453) =

43.28085 - 43.28085*.8908987 =

43.28085 - 38.558853 =

4.721997 grams of "X" after 5 days

That appears to be correct. Did I interpret the formula correctly?
Yes that is correct.

6. Okay, and thanks to all for helping me.
This message board is incredible.